CEMC Euclid, 2013

Beginning with the given equation, we have $$\text{1 x - x amp; = 3 1 x - x x amp; = 3 1- x amp; = 3 x (since x 0 ) (1- x) 2 amp; = 9 2 x (squaring both sides) 1-2 x + 2 x amp; = 9(1- 2 x) 10 2 x - 2 x - 8 amp;= 0 5 2 x - x - 4 amp; = 0 (5 x + 4)( x-1) amp; = 0}$$ Therefore, $\sin x = -\dfrac{4}{5}$ or $\sin x = 1$.

If $\sin x = 1$, then $\cos x = 0$ and $\tan x$ is undefined, which is inadmissible in the original equation.

Therefore, $\sin x = -\dfrac{4}{5}$.

(We can check that if $\sin x = -\dfrac{4}{5}$, then $\cos x = \pm\dfrac{3}{5}$ and the possibility that $\cos x = \dfrac{3}{5}$ satisfies the original equation, since in this case $\dfrac{1}{\cos x} = \dfrac{5}{3}$ and $\tan x = -\dfrac{4}{3}$ and the difference between these fractions is 3.)
Since $f(x)=ax+b$, we can determine an expression for $g(x)=f^{-1}(x)$ by letting $y=f(x)$ to obtain $y=ax+b$. We then interchange $x$ and $y$ to obtain $x=ay+b$ which we solve for $y$ to obtain $ay=x-b$ or $y = \dfrac{x}{a} - \dfrac{b}{a}$.

Therefore, $f^{-1}(x) = \dfrac{x}{a} - \dfrac{b}{a}$.

Note that $a \neq 0$. (This makes sense since the function $f(x)=b$ has a graph which is a horizontal line, and so cannot be invertible.)

Therefore, the equation $f(x)-g(x)=44$ becomes $(ax+b)-\left(\dfrac{x}{a} - \dfrac{b}{a}\right) = 44$ or $\left(a-\dfrac{1}{a}\right)x + \left(b+\dfrac{b}{a}\right) = 44 = 0x+44$, and this equation is true for all $x$.

We can proceed in two ways.

Method #1: Comparing coefficients

Since the equation $$\left(a-\dfrac{1}{a}\right)x + \left(b+\dfrac{b}{a}\right) = 0x+44$$ is true for all $x$, then the coefficients of the linear expression on the left side must match the coefficients of the linear expression on the right side.

Therefore, $a-\dfrac{1}{a} = 0$ and $b+\dfrac{b}{a} = 44$.

From the first of these equations, we obtain $a = \dfrac{1}{a}$ or $a^2=1$, which gives $a=1$ or $a=-1$.

If $a=1$, the equation $b+\dfrac{b}{a} = 44$ becomes $b+b=44$, which gives $b=22$.

If $a=-1$, the equation $b+\dfrac{b}{a} = 44$ becomes $b-b=44$, which is not possible.

Therefore, we must have $a=1$ and $b=22$, and so $f(x)=x+22$.

Method #2: Trying specific values for $x$

Since the equation $$\left(a-\dfrac{1}{a}\right)x + \left(b+\dfrac{b}{a}\right) = 0x+44$$ is true for all values of $x$, then it must be true for any specific values of $x$ that we choose.

Choosing $x=0$, we obtain $0+\left(b+\dfrac{b}{a}\right) = 44$ or $b+\dfrac{b}{a} = 44$.

Choosing $x=b$, we obtain $\left(a-\dfrac{1}{a}\right)b + \left(b+\dfrac{b}{a}\right) = 44$ or $ab+b = 44$.

We can rearrange the first of these equations to get $\dfrac{ab+b}{a}=44$.

Using the second equation, we obtain $\dfrac{44}{a}=44$ or $a=1$.

Since $a=1$, then $ab+b=44$ gives $2b=44$ or $b=22$.

Thus, $f(x)=x+22$.

In summary, the only linear function $f$ for which the given equation is true for all $x$ is $f(x)=x+22$.